Bài 28: Luyện tập: Tính chất của kim loại kiềm, kim loại kiềm thổ và hợp chất của chúng
Hướng dẫn giải Bài 1 (Trang 132 SGK Hóa học 12)
<p>Cho 3,04 gam hỗn hợp NaOH v&agrave; KOH t&aacute;c dụng với axit HCl thu được 4,15 gam hỗn hợp muối clorua. Khối lượng của mỗi hidroxit trong hỗn hợp lần lượt l&agrave;</p> <p>A. 1,17 gam v&agrave; 2,98 gam.&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;C. 1,12 gam v&agrave; 1,6 gam.</p> <p>C. 1,12 gam v&agrave; 1,92 gam.&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;D. 0,8 gam v&agrave; 2,24 gam,</p> <p><strong>Giải</strong></p> <p><em>Chọn D</em>. Đặt số mol NaOH v&agrave; KOH lần lượt l&agrave; x mol v&agrave; y mol.</p> <p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>NaOH</mi><mo>&#160;</mo><mo>+</mo><mo>&#160;</mo><mi>HCl</mi><mo>&#160;</mo><mo>&#8594;</mo><mi>NaCl</mi><mo>&#160;</mo><mo>+</mo><mo>&#160;</mo><msub><mi mathvariant="normal">H</mi><mn>2</mn></msub><mi mathvariant="normal">O</mi><mspace linebreak="newline"/><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mi mathvariant="normal">x</mi><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#8594;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mi mathvariant="normal">x</mi><mspace linebreak="newline"/><mi>KOH</mi><mo>&#160;</mo><mo>+</mo><mo>&#160;</mo><mi>HCl</mi><mo>&#160;</mo><mo>&#8594;</mo><mo>&#160;</mo><mi>KCl</mi><mo>&#160;</mo><mo>+</mo><mo>&#160;</mo><msub><mi mathvariant="normal">H</mi><mn>2</mn></msub><mi mathvariant="normal">O</mi><mspace linebreak="newline"/><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mi mathvariant="normal">y</mi><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#8594;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mi mathvariant="normal">y</mi><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo></math></p> <p>Theo đề b&agrave;i ta c&oacute;:&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="{" close=""><mtable columnalign="left"><mtr><mtd><mn>40</mn><mi mathvariant="normal">x</mi><mo>&#160;</mo><mo>+</mo><mo>&#160;</mo><mn>56</mn><mi mathvariant="normal">y</mi><mo>&#160;</mo><mo>=</mo><mo>&#160;</mo><mn>3</mn><mo>,</mo><mn>04</mn></mtd></mtr><mtr><mtd><mn>58</mn><mo>,</mo><mn>5</mn><mi mathvariant="normal">x</mi><mo>&#160;</mo><mo>+</mo><mo>&#160;</mo><mn>74</mn><mo>,</mo><mn>5</mn><mi mathvariant="normal">y</mi><mo>&#160;</mo><mo>=</mo><mo>&#160;</mo><mn>4</mn><mo>,</mo><mn>15</mn></mtd></mtr></mtable></mfenced><mo>&#8660;</mo><mo>&#160;</mo><mfenced open="{" close=""><mtable columnalign="left"><mtr><mtd><mi mathvariant="normal">x</mi><mo>&#160;</mo><mo>=</mo><mo>&#160;</mo><mn>0</mn><mo>,</mo><mn>02</mn></mtd></mtr><mtr><mtd><mi mathvariant="normal">y</mi><mo>&#160;</mo><mo>=</mo><mo>&#160;</mo><mn>0</mn><mo>,</mo><mn>04</mn></mtd></mtr></mtable></mfenced></math></p> <p>Khối lượng mỗi hidroxit trong hỗn hợp:</p> <p>&nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi mathvariant="normal">m</mi><mi>NaOH</mi></msub><mo>&#160;</mo><mo>=</mo><mo>&#160;</mo><mn>0</mn><mo>,</mo><mn>02</mn><mo>.</mo><mn>40</mn><mo>&#160;</mo><mo>=</mo><mo>&#160;</mo><mn>0</mn><mo>,</mo><mn>8</mn><mo>&#160;</mo><mo>(</mo><mi mathvariant="normal">g</mi><mo>)</mo><mo>;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><mo>&#160;</mo><msub><mi mathvariant="normal">m</mi><mi>KOH</mi></msub><mo>&#160;</mo><mo>=</mo><mo>&#160;</mo><mn>0</mn><mo>,</mo><mn>04</mn><mo>.</mo><mn>56</mn><mo>&#160;</mo><mo>=</mo><mo>&#160;</mo><mn>2</mn><mo>,</mo><mn>24</mn><mo>&#160;</mo><mo>(</mo><mi mathvariant="normal">g</mi><mo>)</mo></math></p> <p>&nbsp;</p>
Hướng dẫn Giải Bài 1 (Trang 132, SGK Hóa học 12)
GV: GV colearn
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Hướng dẫn Giải Bài 1 (Trang 132, SGK Hóa học 12)
GV: GV colearn